Indefinite Integration
Integration of Rational Functions
Grade 12
Question:
<p>[JEE Advanced 2006] \(\displaystyle\int\frac{x}{(x-1)(x^2+1)}\,dx\) equals (where \(C\) is constant)</p>
<li>\(\dfrac12\ln|x-1|+\dfrac12\tan^{-1}x+C\)</li>
<li>\(-\dfrac12\ln(x^2+1)+\dfrac14\tan^{-1}x+C\)</li>
<li>\(\dfrac14\ln|x-1|+\tan^{-1}x+C\)</li>
<li>\(\dfrac12\ln|x-1|-\dfrac14\ln(x^2+1)+\dfrac12\tan^{-1}x+C\)</li>
Step-by-Step Solution
Key Concept: Partial fractions: x/((x-1)(x^2+1)) = A/(x-1) + (Bx+C)/(x^2+1). At x=1: A=1/2. Compare and find B=-1/2, C=1/2.
<p>Partial fractions: \(\dfrac{x}{(x-1)(x^2+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}\).</p>
<p>At \(x=1\): \(1=2A\Rightarrow A=\frac12\).</p>
<p>Compare \(x^2\): \(0=A+B\Rightarrow B=-\frac12\). Constant: \(0=-A+C\Rightarrow C=\frac12\).</p>
<p>\[I = \frac12\int\frac{dx}{x-1}+\int\frac{-\frac12 x+\frac12}{x^2+1}\,dx\]</p>
<p>\[= \frac12\ln|x-1|-\frac14\ln(x^2+1)+\frac12\tan^{-1}x+C\]</p>
<p>Answer: <strong>(D)</strong></p>
Correct Answer: D